(i)[2]
Demonstrate that $y = \dfrac{e^{-x}}{1 + e^{-x}}$.
(ii)[4]
Hence prove that $\int_0^1 y\,dx = \ln\left(\frac{2e}{e + 1}\right)$.
Mathematics 9709 · AS & A Level · Integration
Integration — practice question
You are given $x = \ln(1 - y) - \ln y$, where $0 \le y \le 1$.
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This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Eliminate the logarithms correctly and arrive at $e^x=\frac{1-y}{y}$” …
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